Step 2: Rewrite in Cartesian form:
c= (-2,2), in the complex plane
Step 3: Rewrite in polar form:
We need to rewrite c in the following form: , where the complex number z in this case is c
(i) Find θ by solving θ= arctan(y/x).
Since, as shown in Step 2, y= 2 and x= -2, (y/x)= (2/-2)= -1.
Thus, we have θ= arctan(-1)= 3π/4 for 0≤θ≤2π
(ii) Find r by solving
Since x=2 and y=-2, we have:
r=sqrt{(-2)^2+ (2)^2}
r=sqrt {4 + 4}
r=sqrt {8} (You can simplify this further, but you do not need to)
(iii) Bring it all together:
So, we have c= sqrt{8}(e^(i3π/4)) in polar form
1) (b) Using polar coordinates, find all the cube roots of c. (Hint: A cube root is a complex number such that z^3 = c. How many complex solutions does this equation have?)
If we rewrite z^3= c as z^3 - c = 0, we see that the equation will have exactly three solution (by the Fundamental Theorem of Algebra). If we write c is polar form (which we solved in part (a)), we have c= sqrt{8}(e^(i3π/4)). So, z^3=sqrt{8}(e^(i3π/4))(e^(i2πk) for any integer k.
Now, solve for z by raising both sides to the (1/3)rd power:
z^3 = sqrt{8}(e^(i3π/4))(e^(i2πk))
z^3(1/3) = sqrt{8}(1/3)(e^(i3π/4)(1/3))(e^(i2πk)(1/3))
z = sqrt{2}(e^(iπ/4)(e^(i2πk/3))
To find the three solutions, we first restrict our θ so that 0≤θ≤2π (to avoid repeats), and solve for when k=0,1, and 2 (we stop at two since, after 2, θ becomes greater than 2π and, therefore, cycles).
QUIZ 1 SOLUTIONS
1) a) Express the complex number c=(1+i)^3 in both Cartesian and polar forms.
Step 1: Expand c= (1+i)^3:
c= (1+i)(1+2i+i^2)
c= (1+i)(1+i-1)
c= (1+i)(2i)
c= 2i+2i^2
c= 2i-2
c= -2+2i
Step 2: Rewrite in Cartesian form:
c= (-2,2), in the complex plane
Step 3: Rewrite in polar form:
We need to rewrite c in the following form:
(i) Find θ by solving θ= arctan(y/x).
Since, as shown in Step 2, y= 2 and x= -2, (y/x)= (2/-2)= -1.
Thus, we have θ= arctan(-1)= 3π/4 for 0≤θ≤2π
(ii) Find r by solving
Since x=2 and y=-2, we have:
r=sqrt{(-2)^2+ (2)^2}
r=sqrt {4 + 4}
r=sqrt {8} (You can simplify this further, but you do not need to)
(iii) Bring it all together:
So, we have c= sqrt{8}(e^(i3π/4)) in polar form
1) (b) Using polar coordinates, find all the cube roots of c. (Hint: A cube root is a complex number such that z^3 = c. How many complex solutions does this equation have?)
If we rewrite z^3= c as z^3 - c = 0, we see that the equation will have exactly three solution (by the Fundamental Theorem of Algebra). If we write c is polar form (which we solved in part (a)), we have c= sqrt{8}(e^(i3π/4)). So, z^3=sqrt{8}(e^(i3π/4))(e^(i2πk) for any integer k.
Now, solve for z by raising both sides to the (1/3)rd power:
z^3 = sqrt{8}(e^(i3π/4))(e^(i2πk))
z^3(1/3) = sqrt{8}(1/3)(e^(i3π/4)(1/3))(e^(i2πk)(1/3))
z = sqrt{2}(e^(iπ/4)(e^(i2πk/3))
To find the three solutions, we first restrict our θ so that 0≤θ≤2π (to avoid repeats), and solve for when k=0,1, and 2 (we stop at two since, after 2, θ becomes greater than 2π and, therefore, cycles).
Solve for k=0
z1= sqrt{2}(e^(iπ/4)(e^(0))
z1 = sqrt{2}(e^(iπ/4)(1)
z1 = sqrt{2}(e^(iπ/4)
Solve for k=1
z2 = sqrt{2}(e^(iπ/4))(e^(i2π/3))
z2 = sqrt{2}(e^(i11π/12)
Solve for k=3
z3 = sqrt{2}(e^(iπ/4))(e^(i4π/3))
z3 = sqrt{2}(e^(i19π/12)
The three solutions to z^3= c are z1 = sqrt{2}(e^(iπ/4), z2 = sqrt{2}(e^(i11π/12), and z3 = sqrt{2}(e^(i19π/12)